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how to get Parameters from a json file with python

I am working in Python with a JSON file. When I print date it comes out like this :
print (date['slot'])
[{'slot': 1, 'type': {'url': 'https://pokeapi.co/api/v2/type/11/', 'name':
'water'}}]
But I just want to get 'water'. I have been using different methods and it did not work. Can someone please help me to get just 'water' from the code?

print (date['slot'][0]['type']['name'])
OR
print (date['slot'][0]['type']['name'] if date['slot'] else None)
OR More generic
for sd in date['slot']:
print (sd['type']['name'])

How to extract some data from url using Python

I have an url as follows:
https://some_url/vivi/v2/ZUxOZmVrdzJqTURxV20wQ0RvRld6SytEQWNocThwMGVnbFJ4RDQrZzJMeGRBcnhPYnUzV1pRPT0=/BE?category=PASSENGER&make=30&model=124&regmonth=3&regdate=2015-03&body=443,4781&facelift=252&seats=4&bodyHeight=443&bodyLength=443&weight=-1&engine=1394&wheeldrive=196&transmission=400
What I need is to get the string after v2/, thus ZUxOZmVrdzJqTURxV20wQ0RvRld6SytEQWNocThwMGVnbFJ4RDQrZzJMeGRBcnhPYnUzV1pRPT0=
I use furl to extract the parameter value. I do it as follows:
furl(url).args['category'] // gives PASSENGER
But here I do not have the name of the parameter.
How can I do that?

If you don't need a generalized solution but for the url you have provided in question. Then you can do the following:
url="https://some_url/vivi/v2/ZUxOZmVrdzJqTURxV20wQ0RvRld6SytEQWNocThwMGVnbFJ4RDQrZzJMeGRBcnhPYnUzV1pRPT0=/BE?category=PASSENGER&make=30&model=124&regmonth=3&regdate=2015-03&body=443,4781&facelift=252&seats=4&bodyHeight=443&bodyLength=443&weight=-1&engine=1394&wheeldrive=196&transmission=400"
answer=url.split('/')[5]

Use following code:
l=url.split('/')
m=l[l.index('v2')+1]
print(m)

Desired output using re.
import re
url = "https://some_url/vivi/v2/ZUxOZmVrdzJqTURxV20wQ0RvRld6SytEQWNocThwMGVnbFJ4RDQrZzJMeGRBcnhPYnUzV1pRPT0=/BE?category=PASSENGER&make=30&model=124&regmonth=3&regdate=2015-03&body=443,4781&facelift=252&seats=4&bodyHeight=443&bodyLength=443&weight=-1&engine=1394&wheeldrive=196&transmission=400"
re.findall(r'v2/(.*)/', url)
Resulting with ['ZUxOZmVrdzJqTURxV20wQ0RvRld6SytEQWNocThwMGVnbFJ4RDQrZzJMeGRBcnhPYnUzV1pRPT0='].
But it's safer to use split() the way other mentioned, because when api version changes to v3 this re code won't work anymore.

The string that you are after is not a query parameter, it is part of the URL path.
In the general case you can use the urllib.parse module to parse the URL into its components, then access the path. Then extract the required part of the path:
import base64
from urllib.parse import urlparse, parse_qs
parsed_url = urlparse(url)
s = parsed_url.path.split('/')[-2] # second last component of path
>>> s
'ZUxOZmVrdzJqTURxV20wQ0RvRld6SytEQWNocThwMGVnbFJ4RDQrZzJMeGRBcnhPYnUzV1pRPT0='
>>> base64.b64decode(s)
b'eLNfekw2jMDqWm0CDoFWzK+DAchq8p0eglRxD4+g2LxdArxObu3WZQ=='
The keys and values of the query string can also be processed into a dictionary and accessed by key:
params = parse_qs(parsed_url.query)
>>> params
{'category': ['PASSENGER'], 'make': ['30'], 'model': ['124'], 'regmonth': ['3'], 'regdate': ['2015-03'], 'body': ['443,4781'], 'facelift': ['252'], 'seats': ['4'], 'bodyHeight': ['443'], 'bodyLength': ['443'], 'weight': ['-1'], 'engine': ['1394'], 'wheeldrive': ['196'], 'transmission': ['400']}
>>> params['category']
['PASSENGER']

Handling web links in Python

I have some links stored in a file which looks like this:
http://r14---sn-p5qlsnss.googlevideo.com/videoplayback?itag=22&id=o-AOtM1kWozUiJKP2ENWH989ZIfJaZNPVvXTrBkXx40lG5&key=yt6&ip=159.253.144.86&lmt=1480060612064057&dur=1047.870&mv=m&source=youtube&ms=au&ei=DtN8WLfwFsKb1gKXho6YDw&expire=1484597102&mn=sn-p5qlsnss&mm=31&ipbits=0&nh=IgpwcjAzLmlhZDA3KgkxMjcuMC4wLjE&initcwndbps=4717500&mt=1484575249&pl=24&signature=1ECAB2B56C30CBF760721A1A26A7E80963DB36B8.6336B2C9C41DB53C8FA1D2A037793275F57C4825&ratebypass=yes&mime=video%2Fmp4&upn=tUcEt34Qe6c&sparams=dur%2Cei%2Cid%2Cinitcwndbps%2Cip%2Cipbits%2Citag%2Clmt%2Cmime%2Cmm%2Cmn%2Cms%2Cmv%2Cnh%2Cpl%2Cratebypass%2Csource%2Cupn%2Cexpire&title=600ft+UFO+Crash+Site+Discovered+On+Mars%21+11%2F23%2F16
At the end of the link we have the video's title. I want to read this link from a file and get the video's title in a proper format (with those '+' and '%' signs properly resolved). How do I do that?
I cannot use raw cgi as suggested here since the link is read from a file and not submitted by a form. Any idea?

There's super convenient urllib.parse.parse_qs for python 3, but if you're using python 2, you might have to dig out the title string first.
import urllib
url = 'http://r14---sn-p5qlsnss.googlevideo.com/videoplayback?itag=22&id=o-AOtM1kWozUiJKP2ENWH989ZIfJaZNPVvXTrBkXx40lG5&key=yt6&ip=159.253.144.86&lmt=1480060612064057&dur=1047.870&mv=m&source=youtube&ms=au&ei=DtN8WLfwFsKb1gKXho6YDw&expire=1484597102&mn=sn-p5qlsnss&mm=31&ipbits=0&nh=IgpwcjAzLmlhZDA3KgkxMjcuMC4wLjE&initcwndbps=4717500&mt=1484575249&pl=24&signature=1ECAB2B56C30CBF760721A1A26A7E80963DB36B8.6336B2C9C41DB53C8FA1D2A037793275F57C4825&ratebypass=yes&mime=video%2Fmp4&upn=tUcEt34Qe6c&sparams=dur%2Cei%2Cid%2Cinitcwndbps%2Cip%2Cipbits%2Citag%2Clmt%2Cmime%2Cmm%2Cmn%2Cms%2Cmv%2Cnh%2Cpl%2Cratebypass%2Csource%2Cupn%2Cexpire&title=600ft+UFO+Crash+Site+Discovered+On+Mars%21+11%2F23%2F16'
title = url[url.rfind('&title=') + 7:]
print urllib.unquote_plus(title)
Note: thanks to bereal for pointing out parse_qs is also available in python 2, so just:
import urlparse
print urlparse.parse_qs(url)['title'][0]
'600ft UFO Crash Site Discovered On Mars! 11/23/16'

You could use urllib.parse.parse_qs and give it the string:
In [17]: urllib.parse.parse_qs(s)
Out[17]:
{'dur': ['1047.870'],
'ei': ['DtN8WLfwFsKb1gKXho6YDw'],
'expire': ['1484597102'],
'http://r14---sn-p5qlsnss.googlevideo.com/videoplayback?itag': ['22'],
[.. and so on ..]
'source': ['youtube'],
'sparams': ['dur,ei,id,initcwndbps,ip,ipbits,itag,lmt,mime,mm,mn,ms,mv,nh,pl,ratebypass,source,upn,expire'],
'title': ['600ft UFO Crash Site Discovered On Mars! 11/23/16'],
'upn': ['tUcEt34Qe6c']}
In [18]: urllib.parse.parse_qs(s)["title"][0]
Out[18]: '600ft UFO Crash Site Discovered On Mars! 11/23/16'

Purl can fit your needs:
import purl
u = purl.URL('http://r14---sn-p5qlsnss.googlevideo.com/videoplayback?itag=22&id=o-AOtM1kWozUiJKP2ENWH989ZIfJaZNPVvXTrBkXx40lG5&key=yt6&ip=159.253.144.86&lmt=1480060612064057&dur=1047.870&mv=m&source=youtube&ms=au&ei=DtN8WLfwFsKb1gKXho6YDw&expire=1484597102&mn=sn-p5qlsnss&mm=31&ipbits=0&nh=IgpwcjAzLmlhZDA3KgkxMjcuMC4wLjE&initcwndbps=4717500&mt=1484575249&pl=24&signature=1ECAB2B56C30CBF760721A1A26A7E80963DB36B8.6336B2C9C41DB53C8FA1D2A037793275F57C4825&ratebypass=yes&mime=video%2Fmp4&upn=tUcEt34Qe6c&sparams=dur%2Cei%2Cid%2Cinitcwndbps%2Cip%2Cipbits%2Citag%2Clmt%2Cmime%2Cmm%2Cmn%2Cms%2Cmv%2Cnh%2Cpl%2Cratebypass%2Csource%2Cupn%2Cexpire&title=600ft+UFO+Crash+Site+Discovered+On+Mars%21+11%2F23%2F16')
print(u.query_param('title'))

use urlparse.parse_qs:
try:
from urlparse import urlparse # for python2
except:
from urllib import parse as urlparse # for python3
rv = urlparse.parse_qs(link)
title = rv['title'][0]

import urllib
a = "http://r14---sn-p5qlsnss.googlevideo.com/videoplayback?itag=22&id=o-AOtM1kWozUiJKP2ENWH989ZIfJaZNPVvXTrBkXx40lG5&key=yt6&ip=159.253.144.86&lmt=1480060612064057&dur=1047.870&mv=m&source=youtube&ms=au&ei=DtN8WLfwFsKb1gKXho6YDw&expire=1484597102&mn=sn-p5qlsnss&mm=31&ipbits=0&nh=IgpwcjAzLmlhZDA3KgkxMjcuMC4wLjE&initcwndbps=4717500&mt=1484575249&pl=24&signature=1ECAB2B56C30CBF760721A1A26A7E80963DB36B8.6336B2C9C41DB53C8FA1D2A037793275F57C4825&ratebypass=yes&mime=video%2Fmp4&upn=tUcEt34Qe6c&sparams=dur%2Cei%2Cid%2Cinitcwndbps%2Cip%2Cipbits%2Citag%2Clmt%2Cmime%2Cmm%2Cmn%2Cms%2Cmv%2Cnh%2Cpl%2Cratebypass%2Csource%2Cupn%2Cexpire&title=600ft+UFO+Crash+Site+Discovered+On+Mars%21+11%2F23%2F16"
b = a.split('=')[-1]
print urllib.unquote_plus(b)

Get the description of a status code in Python Requests

I would like to be able to enter a server response code and have Requests tell me what the code means. For example, code 200 --> ok
I found a link to the source code which shows the dictionary structure of the codes and descriptions. I see that Requests will return a response code for a given description:
print requests.codes.processing # returns 102
print requests.codes.ok # returns 200
print requests.codes.not_found # returns 404
But not the other way around:
print requests.codes[200] # returns None
print requests.codes.viewkeys() # returns dict_keys([])
print requests.codes.keys() # returns []
I thought this would be a routine task, but cannot seem to find an answer to this in online searching, or in the documentation.

Alternatively, in case of Python 2.x, you can use httplib.responses:
>>> import httplib
>>> httplib.responses[200]
'OK'
>>> httplib.responses[404]
'Not Found'
In Python 3.x, use http module:
In [1]: from http.client import responses
In [2]: responses[200]
Out[2]: 'OK'
In [3]: responses[404]
Out[3]: 'Not Found'

One possibility:
>>> import requests
>>> requests.status_codes._codes[200]
('ok', 'okay', 'all_ok', 'all_okay', 'all_good', '\\o/', '\xe2\x9c\x93')
The first value in the tuple is used as the conventional code key.

I had the same problem before and found the
answer in this question
Basically:
responsedata.status_code - gives you the integer status code
responsedata.reason - gives the text/string representation of the status code

requests.status_codes.codes.OK
works nicely and makes it more readable in my application code
Notice that in source code: the requests.status_codes.codes is of type LookupDict which overrides method getitem
You could see all the supported keys with - dir(requests.status_codes.codes)
When using in combination with FLASK:
i like use following enum from flask-api plugin
from flask_api import status where i get more descriptive version of HTTP status codes as in -
status.HTTP_200_OK

With Python 3.x this will work
>>> from http import HTTPStatus
>>> HTTPStatus(200).phrase
'OK'

Python error when using urllib.open

When I run this:
import urllib
feed = urllib.urlopen("http://www.yahoo.com")
print feed
I get this output in the interactive window (PythonWin):
<addinfourl at 48213968 whose fp = <socket._fileobject object at 0x02E14070>>
I'm expecting to get the source of the above URL. I know this has worked on other computers (like the ones at school) but this is on my laptop and I'm not sure what the problem is here. Also, I don't understand this error at all. What does it mean? Addinfourl? fp? Please help.

Try this:
print feed.read()
See Python docs here.

urllib.urlopen actually returns a file-like object so to retrieve the contents you will need to use:
import urllib
feed = urllib.urlopen("http://www.yahoo.com")
print feed.read()

In python 3.0:
import urllib
import urllib.request
fh = urllib.request.urlopen(url)
html = fh.read().decode("iso-8859-1")
fh.close()
print (html)